Question

Each item produced by a production system goes through a quality inspection process to be classified as non-defective (ND), partially defective (PD), or totally defective (TD). Over the past several years, category percentages for a certain item produced by this system have stabilized at 85% non-defective (ND), 10% partially defective (PD), and 5% totally defective (TD). The company has purchased a new machine for producing this item. A sample of size 200 items produced by the new machine yielded 180 non-defective (ND), 16 partially defective (PD), and 4 totally defective (TD) items. We are interested to perform a goodness of fit test to determine whether the sample of 200 items produced by the new machine is consistent with the historical distribution of the number of items in each of the three categories. Let PND = probability that an item produced is non-defective (ND) PPD = probability that an item produced is partially defective (PD) PTD = probability that an item produced is totally defective (TD)

1. Refer to Exhibit 1. The null hypothesis ( H0) for the goodness of fit test is: H0: The population of items produced by the new machine follows a multinomial probability distribution with PND = 0.85, PPD = 0.10, and PTD = 0.05 . (True or False)

2. What is the expected frequency for non-defective category?

A. 170 B. 180 C. 85 D. 350

3. What is the chi square test statistic?

A. 7.888 B. 11.747 C. 43.166 D. 4.988

4. What is the critical value of the chi-square obtained from the table(or using MS Excel) for a=0.05(5% level significance)?

A. 12.592 B. 5.991 C. 7.378 D. 10.597

5. What is the p-value obtained from the chi-square table (or using MS Excel)?

A. Less than 0.05 B. More than 0.10 C. More than 0.05 but less than 0.10 D. None of the above

6. What is your conclusion?

A. The population of items produced by the new machine follows a normal probability distribution with a mean of 170 and a standard deviation of 5. B. Reject H0, the population of items produced by the new machine does NOT follow a multinomial probability distribution with PND = 0.85, PPD = 0.10, and PTD = 0.05 . C. Do NOT reject H0; the population of items produced by the new machine follows a multinomial probability distribution with PND = 0.85, PPD = 0.10, and PTD = 0.05 . D. The results of the test are inconclusive.

Answer 1

Answer :

##1. Refer to Exhibit 1. The null hypothesis ( H0) for the goodness of fit test is: H0: The population of items produced by the new machine follows a multinomial probability distribution with PND = 0.85, PPD = 0.10, and PTD = 0.05 . (True or False)

Answer : True

( null hypothesis is equal to the proportion for each category )

## 2. What is the expected frequency for non-defective category?

Answer : A. 170

Expected frequency = N * p = 200 * 0.85 = 170

## 3. What is the chi square test statistic?

Answer : D. 4.988

chi square =  Σ ( Oi - Ei)^2 / Ei

Where Oi : observed frequency , Ei = Expected frequency

here Observed frequency for ND = 180

observed frequency for PD = 16 and

observed frequency for TD = 4

Expected frequency for ND = 0.85 * 200 = 170

Expected frequency for PD = 0.10 * 200 = 20

Expected frequency for TD = 0.05 * 200 = 10

chi square test statistics = [( 180 - 170) ^2 / 170] + [ (16 - 20)^2 / 20 ] + [ ( 4 - 10)^2 / 10 ]

= 0.5882 + 0.80 + 3.6

= 4.988

correct option is D   

## 4. What is the critical value of the chi-square obtained from the table(or using MS Excel) for a=0.05(5% level significance)?

Answer : correct option is B

B. 5.991 ( use statistical table)

degree of feedom = k -1 = 2  

  ##5. What is the p-value obtained from the chi-square table (or using MS Excel)?

Answer : correct option is C

C. More than 0.05 but less than 0.10

= 0.0825 ( use statistical table)

0.05 < p value < 0.10

## 6. What is your conclusion?

Answer : correct option is C

C. Do NOT reject H0; the population of items produced by the new machine follows a multinomial probability distribution with PND = 0.85, PPD = 0.10, and PTD = 0.05 .

( we fail to reject Ho because here p value is greater than alpha 0.05 value using p value approch and chi square test statistics < chi square critical value critical value approach , here both p value as well as critical value approach we fail to reject Ho ie fail to reject Ho )